Codeforces 340D D. Bubble Sort Graph (dp+ segment tree)

Topic Links:

Codeforces 340D

Main topic:

Given a program, is the bubble sort, each time if found adjacent to two number before one is greater than the next one, after the swap position to build the edge, ask the last to get the largest independent set in this figure, this maximum independent set defined as all the points are not adjacent to the largest point of the scale of the collection.

Topic Analysis:

• First we can know that for a[i], it will only and must be smaller than the back side of the building, so we just need to fix the first point, and then find the longest ascending subsequence. (This ensures that no adjacent points are available)
• Defines the state Dp[i] is the length of the longest ascending subsequence ending with the I term.
• The transfer equation is as follows: d P[I]=max{d P[J]+1},(1≤J≤I?1,a[J]<a[I])
• The maximum length of the interval can be recorded by the segment tree, and the maximum size is smaller than a[i], and the final optimized complexity is O(N?Lo g n 2 )
  

AC Code:

#include <iostream>#include <cstdio>#include <algorithm>#include <cstring>#define MAX 100007using namespace STD;intN,a[max];structtree{intL,R,MAXN;} tree[max<<2];voidPUSH_UP (intu) {TREE[U].MAXN = max (tree[u<<1].MAXN, tree[u<<1|1].MAXN);}voidBuild (intU,intLintr) {tree[u].l = l; TREE[U].R = R;intMID = L+r>>1; TREE[U].MAXN =0;if(L = = r)return; Build (u<<1, L, mid); Build (u<<1|1, mid+1, r);}voidUpdate (intU,intXintV) {intL = TREE[U].L;intR = TREE[U].R;intMID = L+r>>1;if(L = = r) {tree[u].maxn = v;return; }if(x > Mid) Update (u<<1|1, x, v);ElseUpdate (u<<1, x, v); PUSH_UP (u);}intQuery (intU,intLeft,intright) {intL = TREE[U].L;intR = TREE[U].R;intMID = L+r>>1;if(left <= l && R <= right)returnTREE[U].MAXN;intRET =0;if(Left <= mid && right >= l) ret = max (ret, query (u<<1, left, right));if(left <= R && right > mid) ret = MAX (ret, query (u<<1|1, left, right));returnRET;}intMain () { while( ~scanf("%d", &n)) { for(inti =1; I <= N; i++)scanf("%d", &a[i]); Build (1,1, n);intx,ans=1; Update (1, a[1] ,1); for(inti =2; I <= N; i++) {x = query (1,1, A[i]); ans = max (ans, x+1); Update (1, A[i], x+1); }printf("%d\n", ans); }}

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Codeforces 340D D. Bubble Sort Graph (dp+ segment tree)